Auditory Mountain
September 08 2015
Individual analyses
Frequency threshold task
The frequency thresholds are calculated from an adaptive procedure where deltaF is decreased when a wrong answer is provided and increased for a right answer.
Detection and identification deltaF
The evolution of the delta F values across trials for the detection and identification tasks is ploted bellow:
Thresholds
Each participant among sessions
The thresholds are calculated from the DeltaF values corresponding to the last 10 reversals. The procedure was repeated two times before training (sessions 1 and 2) and two times after (sessions 3 and 4). The figure bellow shows these thresholds across sessions for the detection and identification conditions:
Each participant among sessions in percent of the first session
Detection threshold in function of identification threshold
For each session
Pre vs. post tests
The threshold scores are the means of the four sessions (two pre-test and two post-test)
First vs. second tests
Auditory mountain task
Here are the plots of scores, accuracy and duration in function of trials. The accuracy is the difference between the choosen tone and the highest tone. The duration is the length in seconds of each trial from the begining of the sound to the answer. The score is a mixed value calculated from accuracy and duration according to this equation:
\[ (\frac{1}{(\frac{duration}{100}+1)^2} + 1) \times 600 \times (\frac{accuracy}{100})^2 \]
Thus, the time score is a multiplier between 1 for a very long trial and 2 for a very short one (few seconds). Thus, the accuracy score can be doubled with speed.
The accuracy score is 600 maximum (the idea was to make a score in order to have the pleasure to cross 1000 points when the performance is good). For accuracy and time, the power function is used to provide the maximum of points near the perfect score and have a lot of difference when it is harder.
UPDATE:
The equation used from va01 (20/07/15) was modified to have a more steeper curve and invite people to be more accurate and avoid the strategy to make a lot of trial quickly and win a lot of points:
\[ (\frac{1}{(\frac{duration}{100}+1)^2} + 1) \times 600 \times (\frac{accuracy}{100})^6 \]
Participant’s scores
Accuracy of trials
Durations of trials
Accuracy in function of duration for each trial
The variability of the results is not associated with duration. Each point is a trial.
Path length in function of duration for each trial
A correlation seems to exists between the length of the displacement on the screen and the duration of the trial
Accuracy in function of frequency of the target tone
The frequencies used in the mountain task are randomly taken from 400 Hz to 2400 Hz.
Comparison of frequency thresholds and accuracy
Global analyses
Frequency threshold task
Mean of participant’s thresholds
The mean thresholds are calculated from all participant’s deltaF for each session (first and second in pre-test and third and fourth in post-test) and condition (detection and identification).
Mean of participant’s thresholds (percent of the first session)
Thresholds are also calculated in percent of the first session deltaF.
Auditory mountain task
Mean accuracy of trials
The number of target crossing in the auditory mountain is the number of time the participant cross the target.
Longitudinal analysis
Two participants (va01 and df22) who did the experiment were invited to do four other sessions (one a day) identical to the the first session (pre-test, training, post-test).
The protocol was also identical but the participants were told that the better one could win 10 euros more and that the total score would be calculated from all the training sessions and all the threshold measures.
Thus, the equation used to calculate the total score take into account the global scores to the mountain task of each session, the mean accuracy of all mountain task sessions and the results to each threshold tasks in this way:
\[ ScoreTotal + (1000 \times meanAccuracy) + \frac{20 000 000}{deltaFIdentification} + \frac{20 000 000}{deltaFDetection} \]
va01:
\[ 349715.27 + (1000 \times 561.79) + \frac{20 000 000}{177.01} + \frac{20 000 000}{81.41} = 1270153.69 \]
df22:
\[ 187007.97 + (1000 \times 529.45) + \frac{20 000 000}{245.43} + \frac{20 000 000}{75.3} = 1063541.81 \]
Frequency threshold task
The frequency threshold tests are done before and after each training session. The dotted vertical lines correspond to each first threshold task.
All Subjects
Mean
All Subjects percent
Mean percent
Auditory mountain task
Participant’s scores
Accuracy of trials
Durations of trials
Auditory mountain as a frequency threshold measure
We calculated the efficiency of the auditory mountain task to give an estimate of the frequency threshold. It represents for each trial and each participant the difference between the whole mean of accuracy and the mean from trial 1 to the one presented. This difference is expressed in standard deviation (calculated with the whole trials of each participant). For instance, the value of the trial 21 for mn15 is around one standard deviation. This means that for this participant, the mean of the 21 first trials is away of the global mean by one standard deviation.
The same thing can be done with the mean among participants for each trial.
Session info
devtools::session_info()## Session info --------------------------------------------------------------## setting value
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